Solving systems of equations

The system of linear equations

( i )

( ii )

is equivalent to the matrix equation AX = B where

because of the definition of matrix multiplication.

To solve AX = B we multiply both sides of the equation by the inverse of A

Since we get that

Note that when we multiplied both sides by the inverse, the inverse was put to the left of each term.

Summary

To solve the matrix equation AX = B use

Example 1

Solve the system

( i ) 3x +5y = 6

( ii ) x + 2y = - 7

using the matrix inverse.

Solution .

This is equivalent to solving the matrix equation AX = B where

In example one of the previous section, we saw that the inverse of A was

Consequently,

or

Thus, x = 47 and y = -27

Example 2

Solve the system using the matrix inverse.

( i ) 7x + 9y = 62

( ii ) -3x + 5y = 62

Solution .

In this case

Using the inverse found in example three of the previous section, we have

Thus, x = -4 and y = 10.

The matrix inverse method can be used in systems with three variables, too.

Example 3

Solve the system using the matrix inverse.

( i) x + 2y + 3z = 1

( ii ) x + 3y + 3z = 1

( iii ) 2x + 8y + 7z = 1

Solution .

We must solve AX = B where

We'll use the equation

From example three in the section on finding inverses, we know that

Hence

Thus, x = 4, y =0, and z = -1

In the section on Gaussian elimination with back substitution we indicated that this method generally involves the fewest number of multiplications. Why then would we want to solve a system as we just did with the inverse method? One reason might be that we might have several problems to solve that involve the same matrix A (coefficeints of the variables) but have different values for B (the constants on the right side of the equations). In this case once the inverse of A has been computed it can be used again with a different B.